On the Galois cohomology of unipotent algebraic groupsover local and global function fields

Q.T. NGUYEN and D.T. NGUYEN

Institut des Hautes Etudes Scientifiques

35, route de Chartres

91440 – Bures-sur-Yvette (France)

Aout 2005

IHES/M/04/58

On the Galois cohomology of unipotentalgebraic groups over local and global function

fields

Nguyen Quoc Thang and Nguyen Duy Tan ∗

Institute of Mathematics, 18 Hoang Quoc Viet, CauGiay10307, Hanoi - Vietnam

andInsitut des Hautes Etudes Scientifiques, Le Bois-Marie, 35 Rue de Chartres,

F-91440 Bures-sur-Yvette, France

Abstract

We discuss some results on the triviality and finiteness for Ga-lois cohomology of connected unipotent groups over local and globalfunction fields, and their relation with the closedness of orbits. As ap-plication, we show that a separable additive polynomial over a globalfield k of characteristic p > 0 in two variables is universal over k ifand only if it is so over all completions kv of k.

AMS Mathematics Subject Classification (2000): Primary 11E72,Secondary 18G50, 20G10

1 Introduction

It is well-known by a result of Rosenlicht [Ro1, 2] that if G is a smoothconnected unipotent algebraic group defined over a perfect field k then thegroup structure of G is very simple : there is a normal series of k-subgroupsof G, with each factor isomorphic over k to the additive group Ga (cf. also

∗Supported in part by F. R. P. V., Abdus Salam I. C. T. P. and I. H. E. S. E-mail :nqthang@math.ac.vn, nqt@ihes.fr

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[Bo]). In particular, its first Galois cohomology, which will be denoted byH1(k, G) := H1(Gal(ks/k), G(ks)), where Gal(ks/k) denotes the absolute Ga-lois group of k, is trivial (see. e. g. [Se1]). However this is no longer trueif k is non-perfect. First Rosenlicht [Ro1, 2] and then Tits [Ti] were able toshow certain unusual properties of unipotent groups over non-perfect fields.In [Se1], Chap. III, Serre constructed certain function field k and unipotentk-group G, such that the first Galois cohomology H1(k,G) of G is non-trivial(cf. also Raynaud [SGA 3, Exp. 17]). Furthermore, Oesterle developedin [Oe] a comprehensive arithmetic theory of unipotent groups over globalfields and, in particular, showed that the nature of unipotent groups overnon-perfect fields is quite non-trivial. One example given there showed thatthis first cohomology group may be even infinite for some locally compactfield of characteristic p > 0 (a completion of a global function field). In gen-eral, there are many open questions regarding the arithmetic and geometryof unipotent algebraic groups over non-perfect fields. In particular it is in-teresting to inquire about the Galois cohomology of unipotent groups in thiscase. Here we are interested in the case where k is a global function field ofcharacteristic p > 0. We assume that, unless otherwise stated, all algebraicgroups considered here are linear, thus absolutely reduced (i.e., smooth) asin [Bo], [Oe]. We proceed in this paper using some main results from thetheory of unipotent groups over non-perfect fields [Oe], [Ro1,2], [Ti] to inves-tigate certain finiteness properties of Galois cohomology of unipotent groupsover non-perfect fields and also related local - global principles over globalfunction fields.

Theorem. (cf. Theorem 3.1) Let k be a global field and let G be a connectedsmooth unipotent group defined over k. Then the first Galois cohomologygroup H1(k,G) is trivial if and only if it is finite and there is a k-embeddingof G into a semisimple (resp. unipotent) smooth simply connected k-group Hsuch that with natural action of H on H/G, the orbits of H(k) are closed in(H/G)(k) in the A(S)-topology of (H/G)(k) for any finite set S of valuationsof k.Theorem. (cf. Theorem 4.8) Let k be a non-perfect field of characteristicp. Let G be a non-trivial (not necessarily smooth) commutative unipotentk-group scheme of dimension ≤ 1, the connected component G◦ of which isnot k-isomorphic to an extension of Ga by an infinitesimal k-group scheme.Assume in addition that if p = 2 then G◦ is neither k-isomorphic to an ex-tension of Ga, nor of the subgroup defined as in Proposition 4.2, b), by an

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infinitesimal group scheme. Then the flat cohomology H1fl(k,G) is infinite.

In particular, for any global field k of positive characteristic and for any non-trivial commutative k-unipotent group scheme G of dimension ≤ 1 with G◦

not k-isomorphic to an extension of Ga, by an infinitesimal k-group scheme,the cohomology set H1

fl(k,G) is infinite.

Theorem. (cf. Theorem 5.1, Proposition 5.4) (Local - global principle)Let k be a global function field and let G be a connected smooth unipotentgroup defined over k of dimension n.a) If n =1, then H1(k, G) is trivial if and only if for all valuations v of k andthe completion kv of k at v, H1(kv, G) is trivial.b) If n > 1, then Then H1(k,G) is trivial if and only if for all valuationsv of k and the completion kv of k at v, H1(kv, G) is trivial and there is anembedding of G into a smooth k-split unipotent group H such that with nat-ural action of H on H/G, the orbits of H(k) are closed in (H/G)(k) in theA(S)-topology of (H/G)(k) for any finite set S of valuations of k.

2 Some definitions and notations

2.1. Recall that a unipotent group G defined over a filed k is called k-woundif any k-morphism of varieties A1 → G is constant. We refer to [Ti], [KMT],[Oe] for detailed discussion of the theory of k-wound groups.

2.2. For a finite field extension K/k and V an affine K-variety, we denoteby RK/k(V ) the restriction of scalars from K to k. (In the case of separableextension, it is just Weil’s restriction of scalars and in the general case, tak-ing into account also the non-separable extension, the definition was givene. g. in [Gr], [DG] or in [Oe], Appendix 3.) For a field k we denote by ks

its separable closure in an algebraic closure k of k, Γ = Gal(ks/k) denotesthe Galois group of ks over k, and for a (smooth) linear algebraic k-group G,H1(k, G) := H1(Γ, G(ks)) denotes the first Galois cohomology set of G.

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3 Triviality of cohomology and closedness of

orbits

Let k be a global field. For a valuation v of k denote by Ov the ring ofv-integers of kv, and by A the adele ring of k. Let S be a non-empty finiteset of valuations of k, A(S) be the ring of adeles of k with trivial componentsbelonging to S. Then we have for any affine k-variety V ,

V (A) =∏v∈S

V (kv)× V (A(S)),

and let pr2 : V (A) → V (A(S)) be the projection on the second factor. Weendow A(S) with the induced topology from the ring of adeles A of k. Theinduced (from A and A(S)) topologies on V (A) and V (A(S)) are called adeleand A(S)-topology, respectively. Let G be an affine k-group scheme. We saythat G satisfies property (∗) in dimension r if the following holds

(∗) For any finite set S of non-equivalent valuations of k, the localizationmap Hr

fl(k,G) → ∏v∈S Hr

fl(kv, G) is surjective

where Hrfl denotes the flat cohomology group in dimension r, whenever it

makes sense (see [Ca], [Mi1,2], [SGA4], [Sh]).We will see later on that many affine group schemes enjoy this property.

3.1. Theorem. Let k be a global field and let G be a connected smoothaffine group defined over k.

a) Assume that G satisfies property (∗). Then the first Galois cohomologygroup H1(k,G) is trivial if and only if it is finite and for some (thus any)embedding of G into the k-group H = SLn, such that with natural action of Hon H/G, the orbits of H(k) are closed in (H/G)(k) in the A(S)-topology of(H/G)(k) for any finite set S containing some fixed finite set S0 of valuationsof k.

b) Assume that G is smooth and unipotent. Then the assertion a) holdswhere H is either smooth semisimple simply connected k-group or smoothunipotent k-split group.

c) Assume that G is semisimple and either k is a number field or k is a

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global function field and the fundamental group of G satisfies the condition(∗) in dimension 1. Then the assertion a) holds for semisimple groups G.

Proof. a) (”Only if” part.) We take a matrix k-linearization ϕ : G ↪→ GLn

such that G ↪→ H := SLn, where GLn (resp. SLn) denotes the general (resp.special) linear group of n×n-matrices over k. Then we know that L = H/Gis smooth over k and we may consider the following exact sequence

1 → G → H → L → 1.

Since H has trivial 1-Galois cohomology (see e.g. [Se1]), so H1(k,H) = 0,and H1(kv, H) = 0, and from [Oe], Chap. III, it follows that H1(Γ, H(A(S)⊗ks)) = 0. Since H, G are smooth algebraic groups over k, they are also smoothas A-group schemes and as A(S)-group schemes. It follows from [SGA 4] (cf.also [Mi1], Chap. III) that we have canonical isomorphisms between etalecohomology and Galois cohomology

Hiet(A(S), G) ' Hi(Γ, G(A(S)⊗ ks)),

Hiet(A(S), H) ' Hi(Γ, H(A(S)⊗ ks)),

Hiet(A, G) ' Hi(Γ, G(A⊗ ks)),

Hiet(A, H) ' Hi(Γ, H(A⊗ ks)),

(i = 0, 1) and we have the following commutative diagrams with exact rows

G(k)i→ H(k)

r→ L(k)δ→ H1(k,G) → 0

↓ γ ↓ β ↓ α ↓

G(A)iA→ H(A)

rA→ L(A)δA→ H1(Γ, G(A⊗ ks)) → 0

and

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G(k)i→ H(k)

r→ L(k)δ→ H1(k,G) → 0

↓ γS ↓ βS ↓ αS ↓

G(A(S))iS→ H(A(S))

rS→ L(A(S))δS→ H1(Γ, G(A(S)⊗ ks)) → 0

where the maps δ, δA, δS are surjective. By assumption the localizationmaps are surjective, so from the triviality of H1(k, G) it follows the trivialityof H1(kv, G) for all v. Thus from [Oe, Ch. I] and from above it follows thatH1(Γ, G(A(S)⊗ ks)) and H1(Γ, G(A⊗ ks)) are trivial, so r and rS are surjec-tive. From this the assertion follows.

(”If” part.) Assume that G ↪→ SLn for some n. We proceed to prove thefollowing lemmas.

3.2. Lemma. With the notation and assumption in the theorem, for somefinite set of valuations S0 and any finite set S containing S0,the map αS isinjective.

Proof. Since H is a semisimple simply connected k-group, it is well-knownthat there exists a finite set S0 of valuations such that with respect to S0,H has the strong approximation in the global field k, i.e., for any finiteset S containing S0, the image γS(H(k)) is dense in H(A(S)) in the adeletopology. We may endow H1(k,G) and H1(Γ, G(A(S) ⊗ ks)) with weak-est topology such that all maps in the above diagram be continuous. Herethe sets G(k), H(k) and L(k) are endowed with the topology induced fromG(A(S)), H(A(S)), L(A(S)), respectively.

Let x ∈ Ker (αS) and let y ∈ L(k) such that δ(y) = x. Then δS(βS(y)) =0, i.e.,

βS(y) ∈ rS(H(A(S))) = Cl(βS(r(H(k))).

Since the adele topology has a countable basis of topology, there is a se-quence hn ∈ H(k) such that βS(r(hn)) → βS(y) in A(S)-topology. Since βS

is just the diagonal embedding L(k) ↪→ L(A(S)), it follows that r(hn) → yin L(k) in the topology induced from A(S))-topology on L(A(S)). Henceδ(r(hn)) → δ(y) in H1(k, G). Since H(k)-orbits are closed in L(k) with re-spect to the A(S)-topology induced on L(k), and they are finite in number,it follows that all they are open. They are precisely the fibers over elements

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of H1(k, G). Therefore for sufficiently large n we have δ(r(hn)) = δ(y), sox = 0, thus αS is injective. The lemma is proved.

Next we need the following lemma.

3.3. Lemma. a) Assume that in the following commutative diagram ofpointed topological spaces,

Ap→ B

q→ C → 1

α′ ↓ ↓ β′ ↓ γ′

A′ p′→ B′ q′→ C ′ → 1

where A, A’ are topological groups and all maps are continuous and all rowsare exact. If Im (β′) is dense in B’ then so is Im (γ′) in C’.b) If A, A’, B, B’, C and C’ are commutative topological groups, Im (α′) isdense in A′ and the closure of the image of β′ is open in B′, then we havethe following topological isomorphism

B′/Cl(Im (β′)) ' C ′/Cl(Im (γ′)).

Proof. a) Trivial.

b) We have an obvious surjective homomorphism q′ : B′ → C ′/Cl(Im (γ′)).Next we show that q′ induces a surjective homomorphism

q : B′/Cl(Im (β′)) ' C ′/Cl(Im (γ′)).

For this it suffices to show that

q′(Cl(β′(B))) ⊂ Cl(q′(β′(B)))(⊂ Cl(Im (γ′))).

Let b′ ∈ Cl(β′(B)), V an arbitrary open neighborhood of q′(b′) in C ′. Weneed to show that V ∩ q′(β′(B))) 6= ∅. Since q′ is continuous, q′−1(V ) is openin B′, which contains b′. Hence q′−1(V )∩β′(B) 6= ∅. Let y ∈ q′−1(V )∩β′(B),so q′(y) ∈ V ∩ q′(β′(B)), so V ∩ q′(β′(B))) 6= ∅ as required.

Next we show that Ker (q) = 0. It is clear that this amounts to provingthat

q′−1(Cl(γ′(C)) = Cl(β′(B)).

Clearly the set on the left contains the one on the right, since q′−1(Cl(γ′(C)))is closed and containing Im (β′)). Notice that

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q′−1(Cl(γ′(C)) = q′−1(Cl(γ′(β′(B))))

= q′−1(Cl(q′(β′(B)))),

hence it suffices to show that Cl(q′(β′(B))) ⊂ q′(Cl(β′(B))). Indeed, assum-ing that this inclusion holds. Then one has

q′−1(Cl(q′(Im β′))) ⊂ q′−1(q′(Cl(Im β′)))

= p′(A′)Cl(Im β′).

Since Im α′ is dense in A′ so we have

p′(A′) = p′(Cl(α′(A)))

⊂ Cl(p′(α′(A)))

= Cl(β′(p(A)))

⊂ Cl(β′(B)),

hence we have q′(Cl(β′(B))) ⊂ Cl(q′(β′(B))) as claimed. Since Cl(β′(B)) isopen in B′ and q′ is an open map (as quotient homomorphism), it followsthat q′(Cl(β′(B))) is an open subgroup of C ′, hence also a closed subgroup.It contains q′(β′(B)), hence also Cl(q′(β′(B))) as required. The lemma isproved.

Next we prove that the image of H1(k,G) in H1(Γ, G(A(S) ⊗ ks)) is triv-ial for S sufficiently large. First we introduce some notations. If char.k = 0,let O be the ring of integers of k, C = SpecO, and if char.k > 0, let k = F(C),which is the function field of a geometrically integral smooth curve C over afinite field F. One may find an affine group scheme GC′ of finite type over anopen affine subset C ′ ⊂ C, such that G is k-isomorphic to the generic fiber ofGC′ . The following argument is in fact a ”folklore”, which follows from resultsof [EGA IV] (cf [Oe]). It is well-known that the subset C ′ ⊂ C of all elementsc ∈ C where the canonical morphism GC → C is smooth, is open (in thiscase we just need Zariski’s result [EGA IV], 0.22.6.8 and 17.5.1). By passingto an affine open subset C ′′ ⊂ C ′ and use [EGA IV], 17.8.2, we see that allfibers over s ∈ C ′′ are smooth. Now GC′′ → C ′′ is smooth and of finite typeand also finite presentation. By [EGA IV], 9.7.8, over a suitable open and

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non-empty subset C ′′′ ⊂ C′′, all fibers will be also geometrically connected,since by assumption G is connected. Thus over C ′′′, all fibers are smooth andconnected group schemes.

Thus, after shrinking C ′ we may assume that all fibers of GC′ are smooth,connected affine groups. Every G-torsor P over k (i.e. principal homogeneousspace of G) comes from certain GC′-torsor PC′ for suitable C ′ (depending onP ). Since H1(k,G) is finite by assumption, we may take C ′ sufficiently small,so that for such C ′, every G-torsor P comes from a GC′-torsor PC′ . For everyclosed point s ∈ C ′ let v(s) be the corresponding valuation with the localring Ov(s), the maximal ideal mv(s), the completion kv(s) (of k at v(s)) andthe finite residue field κ(v(s)). We have the following commutative diagram

H1et(C ′,GC′) → H1

et(Spec k,GC′ ×C′ Spec k)

↓ ↓

H1et(Spec Ov(s),GC′ ×C′ Spec Ov(s)) → H1

et(Spec kv(s),GC′ ×C′ Spec kv(s))

↓ f

H1et(Spec κ(v(s)),GC′ ×C′ Spec κ(v(s)))

and for a GC′-torsor PC′ we have

PC′ 7→ P := PC′ ×C′ Spec k

↓ ↓

PC′ ×C′ Spec Ov(s) 7→ Pkv(s):= P ×Spec k Spec kv(s)

↓

PC′ ×C′ Spec κ(v(s))

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which shows that Pkv(s)is isomorphic to the generic fiber of PC′×C′Spec Ov(s).

Now every closed fiber of GC′ is a smooth and connected algebraic group anddefined over a finite field, hence by Lang’s Theorem (see [La]), it has trivialGalois cohomology. Therefore every closed fiber PC′ ×C′ Spec κ(v(s)) of GC′-torsor PC′ is a trivial torsor. Since GC′ is smooth, from ”Hensel Lemma” (seeGrothendieck [SGA 3, Expose. XXIV, Prop. 8.1, (ii)]) it follows that f is anisomorphism, so PC′ ×C′ Spec Ov(s) (hence also Pkv(s)

) is trivial. From [Oe],Chap. III, Sec. 2.4, it follows that

H1(Γ, G(A(S)⊗ ks)) '∐v 6∈S

H1(kv, G).

Therefore by taking C\C ′ (hence also S = {v(s)|s ∈ C\C ′}) sufficiently large,we see that the image of H1(k,G) in H1(Γ, G(A(S) ⊗ ks)) is trivial and theassertion a) of the theorem follows.

b) If char.k = 0, then the first Galois cohomology of unipotent groups istrivial, and ”only if” part needs proving, but this follows from above. Nowwe assume that char.k > 0. Smooth unipotent groups over any field satisfythe property (∗) by [TT], so the assertion related with semisimple simplyconnected groups is coved in part a), and the rest follows from the fact thatk-split unipotent groups have strong approximation over k for any S.(In the case H is a unipotent group, we have a shorter argument as follows.Let H be a k-split unipotent group. The ”only if” part is proved like above.Conversely, we show that H1(k,G) is trivial. Since H is unipotent and k-split, the quotient space L is k-isomorphic to the affine space Ar for somer (see [Ro1],[Ro 2]). Hence L has strong approximation, thus the closureCl(βS(L(k))) is equal to L(A(S)). Therefore αS is surjective by Lemma 3.3.Since H1(k, G) is finite, it follows that H1(Γ, G(A(S) ⊗ ks)) is finite, too.From [Oe], Chap. III, Sec. 2.4, it follows that

H1(Γ, G(A(S)⊗ ks)) '∐v 6∈S

H1(kv, G),

hence for all v 6∈ S, H1(kv, G) is finite and for almost all v, H1(kv, G) is trivial.Let S0 := {v|H1(kv, G) 6= 0}. Then S0 is finite and for S = S0, according toLemma 3.2, we have an injection

H1(k,G) → H1(Γ, G(A(S)⊗ ks)) '∐v 6∈S

H1(kv, G) = {0},

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thus H1(k, G) is trivial, too.

c) Now we assume that k is a number field. Then Borel and Harder haveproved ([BH], Theorem 1.7), that G satisfies the property (∗). We give herea modification of their proof. Consider the k-covering of G, i.e., a semisimplesimply connected k-group H together with the following exact sequence

1 → F → H → G → 1,

where F is a finite diagonalizable k-subgroup of H. We have the followingcommutative diagram, where each row is an exact sequence of Galois coho-mology

H1(k, F ) → H1(k,H)π→ H1(k, G)

∆→ H2(k, F )

↓ ↓ α ↓ β ↓ γ

∏v∈S H1(kv, F ) → ∏

v∈S H1(kv, H)πS→ ∏

v∈S H1(kv, G)∆S→ ∏

v∈S H2(kv, F )

We take any sufficiently large finite set S of (non-equivalent) valuations of ksuch that S contains the set of all archimedean valuations ∞ of k. Then thewell-known Hasse principle for simply connected groups (see [PR], for thecase of number fields, and [Ha], for the case of function field) tells us that αis bijective, and a result of [BH], Proposition 1.6, says that γ is surjective.Also, classical results due to Kneser [Kn] (for the case of number field), and[T] (for the case of function field), say that ∆ and ∆S are surjective. Thus bychasing on the above diagram we see that for any given gS ∈

∏v∈S H1(kv, G)

there is an element g ∈ H1(k,G) such that

∆S(gS) = ∆S(β(g))).

By twisting the exact sequence

1 → F → H → G → 1

with a cocycle (xs) representing the class g we get another exact sequence

1 → F → xH → xG → 1.

By considering the commutative diagram with rows being exact sequencesof Galois cohomology deduced from this exact sequence as above, we are

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reduced to the case g = 0 (the distinguished element in H1(k,G). Then∆S(gS) = 0, i.e., gS = πS(hS) for some element hS ∈

∏v∈S H1(kv, H), hence,

by using the surjectivity of α, we deduce that gS ∈ Im (β) as required.In the case of global function fields, by using the fact that H1 of simply

connected groups is trivial (Harder’ Theorem [Ha])

H1fl(k,G) → H2

fl(k, F )

is a bijection ([T]), the condition (∗) in dimension 2 for F then means thatthe same holds for G in dimension 1.

3.4. Corollary. Let G be a smooth unipotent group defined over a globalfield k. Then for any non-empty finite set S of non-equivalent valuations ofk, the restriction map

H1(k,G) →∐v 6∈S

H1(kv, G)

is surjective.

Proof. We may embed G into a smooth k-split unipotent group H andthen consider the exact sequence

1 → G → H → L → 1,

and the related diagram as in Lemma 3.2. We now apply Lemma 3.3, noticingthat L has strong approximation over k with respect to any finite non-emptyS.

4 Finiteness for Galois cohomology of unipo-

tent groups

In this section we prove some criteria for the finiteness of Galois cohomologyof unipotent groups over non-perfect fields. A special attention is devotedto the case of local and global function fields. The following simple lemmawhich is well-known [We] but we give here another short proof.

4.1. Lemma. Let k be a global (local) function field of characteristic p > 0.

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Then [k : kp] = p.

Proof. We claim that if k is a field of char p and [k : kp] = p then [k′ : k′p] = p,for all finite extensions k′ over k. If {x1, . . . , xn} is a basis of k′ over k then{xp

1, . . . , xpn} is a basis of k′p over kp. Therefore, [k′ : k] = [k′p : kp], and we

have

[k′ : k′p] =

[k′ : k][k : kp]

k′p : kp= [k : kp].

We first prove the lemma for the case k = Fq(t), t is transcendental over Fq.Let P (T ) be an arbitrary polynomial in Fq[T ]. Then we can write P (T ) =p−1∑i=0

T iP pi (T ). An element a ∈ k is always of form a =

P (t)

Q(t), P (T ), Q(T ) ∈

Fq[T ]. We have

a =P (t)

Q(t)=

P (t)Q(t)p−1

Qp(t)=

p−1∑i=0

ti(

Ri(t)

Q(t)

)p

.

So {1, t, . . . , tp−1} generates k over kp (as a vector over kp). Since t is tran-scendental over Fq, {1, t, . . . , tp−1} is linearly independent over kp. Hence,{1, t, . . . , tp−1} is a basis of k over kp, and [k : kp] = p.

So from above it follows that the lemma is true for global fields.

Now, consider the case of local fields of characteristic p > 0. We may as-sume k = Fq((t)). Let a =

∑i≥i0

aiti, ai ∈ Fq be an element of k. Since every

elements of Fq is a p-power, we can write ai = bpi , bi ∈ Fq. Therefore, we get

a =∑i≥i0

bpi t

i

=p−1∑r=0

∑i≥(i0−r)/p

bppi+rt

pi+r

=p−1∑r=0

tr

∑i≥(i0−r)/p

bpi+rtpi+r

p

=p−1∑r=0

trapr,

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where ar ∈ k. Hence, {1, t, . . . , tp−1} generates k over kp. And 1, t, . . . , tp−1

are linearly independent over kp, so {1, t, . . . , tp−1} is a basis of k over kp and[k : kp] = p.

Next we examine a class of unipotent groups of special type over non-perfect fields of characteristic p > 0, namely the groups defined by the fol-lowing equation

xp0 + txp

1 + · · ·+ tp−1xpp−1 − xp−1 = 0.

This class of groups was first considered by Rosenlicht in [Ro1,2], and thenin [KMT], [Oe], [Ti], where they served as simple but important examples,which will be needed in the sequel. First we consider the case of character-istic 2.

4.2. Proposition. a) Let kv be Fq((t)), q = 2n, G the Fp(t)-subgroupof G2

a defined by the equation y2 = x + tx2. Then |H1(kv, G)| = 2.b) Let k be a global function field of characteristic 2, t an element of k − k2.Let G be the closed subgroup of G2

a defined as above. For v a nontrivial valua-tion of k, let kv be the completion of k with respect to v. Then |H1(kv, G)| = 2,and H1(k,G) is infinite.

Proof. Let P (x, y) = y2 + x + tx2, and consider P as a homomorphismfrom G2

a to Ga. We have an exact sequence of unipotent k-groups

1 → G → G2a

P→ Ga → 1.

From this we have H1(kv, G) ' kv/P (kv × kv) since H1(kv,Ga) = 0.

a) First, we show that

(1) ct−i ∈ P (kv × kv), for all c ∈ Fq, i ≥ 2,

by induction on i. In fact, if c = α2, where α ∈ Fq, then ct−2i = P (0, αt−i).Since any element of Fq is a square, it follows that the statement holds forall even numbers i. Assume that (1) holds for all numbers j with 2 ≤ j ≤ i.If i is odd, then i + 1 is even and from above we know that (1) holds fori + 1, so we are done. If i is even ≥ 2, i = 2j, then 2 ≤ j < i − 1 and byinduction assumption, there are a, b ∈ kv such that αt−j−1 = P (a, b) where

14

c = α2, with α ∈ Fq, and a, b ∈ k. Then we have ct−2j−1 = P (a + αt−j−1, b).Hence (1) also holds for i + 1 = 2j + 1.

Next, we prove that

(2) Fq[[t]] ⊂ P (kv × kv),, i.e.∑i≥0

citi ∈ P (kv × kv),∀ci ∈ Fq.

In fact, if we can find y ∈ Fq[[t]] such that

(3) y2 + dy/dt =∑i≥0

c2i+1t2i,

where (d/dt) denotes the differentiation with respect to t, then

s = ty2 + y +∑i≥0

citi = x2,

for some x ∈ Fq[[t]], since ds/dt = 0. Now, by writing y =∑i≥0

yiti and by

substituting it into the left hand side of the equation (3), we have

y2 + dy/dt =∑i≥0

y2i t

2i +∑i≥0

y2i+1t2i

=∑i≥0

(y2i + y2i+1)t

2i.

By comparing those coefficients with the same power of t in (3) we derive

y2i + y2i+1 = c2i+1,∀i ≥ 0

and by setting y2i = 1, y2i+1 = c2i+1 + y2i , we get a solution for (3).

Finally, we claim that

(4) for any α ∈ Fq, then αt−1 ∈ P (kv × kv) if and only if α = u2 + u, forsome u ∈ Fq.

In fact, if αt−1 = b2 + a + ta2, for some a, b ∈ k, then by differentiatingboth sides with respect to t, we get αt−2 = da/dt+ a2. It suffices to considerthe case when α 6= 0. In this case, a 6= 0, and we can write a = tlu, wherel ∈ Z, u ∈ k, vt(u) = 0, and vt denotes the valuation corresponding to t. Then

15

(5) αt−2 = ltl−1u + tldu/dt + t2lu2.

We havevt(lt

l−1u + tldu/dt + t2lu2) ≥ 0

if l ≥ 0, andvt(lt

l−1u + tldu/dt + t2lu2) = 2l ≤ −4

if l ≤ −2. Therefore, by comparing both sides of (5), we see that l equals−1, and by writing u = u0 + tu1, u0 ∈ Fq, u1 ∈ Fq[[t]], we get α = u2

0 + u0.Evidently, if α = u2 + u, for some u ∈ Fq, then αt−1 = P (ut−1, 0).

From (1), (2), (4) and by observing that the map u 7→ u2 + u is ahomomorphism from (Fq, +) to (Fq, +) with kernel F2, we conclude that|H1(kv, G)| = 2 and the assertion is proved.

b) Let π be a prime (uniformizing element) for v. For simplicity, we identifykv with the field k(v)((π)), where k(v) is the residue field of k respect tov, so k(v) = Fq, with q = 2n. By Lemma 4.1, we can write t = a2 + πb2,a, b ∈ kv. It is easy to check that b 6= 0. By making the change of variablesy′ = b(y + ax), x′ = b2x, one checks that G is isomorphic to the subgroup

G′ = {y′2 = x′ + πx′2}.

Therefore from part a) it follows that |H1(kv, G)| = |H1(k(v)((π)), G′)| = 2.We know that the localization map

H1(k,G) →∏v∈F

H1(kv, G)

is surjective for any finite set F of non-equivalent discrete valuations ([TT]).Since there are infinitely many v, we conclude that H1(k, G) is infinite.

For the case p > 2, we have the following result. The idea of the proofis similar to the above one, but the computation is more complicated, so wegive a full proof.

4.3. Proposition. a) Let k = Fq(t), v be the valuation correspondingto t and kv = Fq((t)), q = pn, p > 2. Let G be the Fp(t)-subgroup of Gp

a de-fined by the equation xp

0 + txp1 + · · ·+ tp−1xp

p−1 = xp−1. Then |H1(kv, G)| = p.b) Let k be a global function field of characteristic p > 2, t an element in

16

k − kp. Let G be the subgroup of Gpa defined as above. Then for infinitely

many non-equivalent valuations v of k, H1(kv, G) 6= {1} and H1(k,G) is in-finite.c) Let kv = Fq((t)), q = pn, p > 2. Let G be the Fp(t)-subgroup of Gp

a definedby the equation xp

0 + txp1 + · · · + tp−1xp

p−1 + xp−1 = 0. Then |H1(kv, G)| = 1(resp. p) if n is odd (resp. even).

Proof. a) We first show that for

P (x0, ..., xp−1) = xp0 + txp

1 + · · ·+ tp−1xpp−1 − xp−1

then

(1) ct−i ∈ P (kv × · · · × kv), for all c ∈ Fq, i ≥ 2,

by induction on i. Since any element of Fq is a p-power, we may writec = βp, β ∈ Fq. We have, for 0 ≤ r ≤ p− 1, r 6= 1,

ct−(pl+r) = P (0, . . . , βt−l−1, . . . , 0),

where all coordinates are 0 except at xp−r. Assume that (1) holds for allnumbers j, with 2 ≤ j ≤ i. If i 6≡ 0 (mod p), then i + 1 6≡ 1 (mod p), soi + 1 = pl + r, where 0 ≤ r ≤ p − 1, r 6= 1, so (1) holds for i + 1, and weare done. If i = pj then 2 ≤ j < i and by induction assumption, there area0, . . . , ap−1 ∈ kv such that βt−(j+1) = P (a0, . . . , ap−1). By setting

x0 = a0, . . . , xp−2 = ap−2, xp−1 = ap−1 + βt−(j+1),

we have

xp0 + txp

1 + · · ·+ tp−1xp−1 − xp−1 = (ap0 + tap

1 + · · ·+ tp−1app−1 − ap−1)

+tp−1(βt−(j+1))p − βt−(j+1)

= (P (a0, ..., ap−1)− βt−(j+1)) + (βpt−(pj+1))

= ct−(pj+1).

Next, we prove that

17

(2) Fq[[t]] ⊂ P (Fq[[t]]× · · · × Fq[[t]]), i.e., for all ci ∈ Fq we have

c(t) :=∑i≥0

citi ∈ P (Fq[[t]]× · · · × Fq[[t]]).

In fact, by differentiating with respect to t both sides of the following equa-tion ∑

i≥0

citi = xp

0 + txp1 + · · ·+ tp−1xp

p−1 − xp−1

(p− 1) times, and by observing that (p− 1)! ≡ −1 (mod p) (Wilson’s the-orem), we get

(3) dp−1c(t)/dtp−1 =∑i≥0

cpi+p−1tpi = xp

p−1 − dp−1xp−1/dtp−1.

First we find xp−1 satisfying (3) as follows. Let xp−1 =∑i≥0

yiti then

xpp−1 − dp−1xp−1/dtp−1 =

∑i≥0

(ypi − ypi+p−1)t

pi.

Using this and by comparing those coefficients with the same power of t in(3), we get

ypi − ypi+p−1 = cpi+p−1,∀i ≥ 0.

By setting ypi+r = 0, 0 ≤ r ≤ p− 2, ypi+p−1 = ypi − cpi+p−1, we get a solution

xp−1 for (3). From this it follows that

d

dt

[(p− 1)!txp

p−1 − dp−2xp−1/dtp−2 − dp−2c(t)/dp−2t]

= 0,

so there exists xp−2 ∈ Fq[[t]] such that

dp−2c(t)/dtp−2 = (p− 2)!xpp−2 + (p− 1)!xp

p−1 − dp−2xp−1/dtp−2.

By repeating this argument p − 2 times, we can find x0, . . . , xp−1 ∈ Fq[[t]]such that c(t) = P (x0, . . . , xp−1), and (2) is proved.

Finally, we claim that

(4) for an element c ∈ Fq, ct−1 ∈ P (kv × · · · × kv) if and only if c = up − u,for some u ∈ Fq.

18

In fact, if ct−1 = P (a0, . . . , ap−1), a0, . . . , ap−1 ∈ kv, then again, by differ-entiating with respect to t this equality (p− 1) times , we get

(5) ct−p = app−1 + dp−1ap−1/dtp−1.

It suffices to consider the case c 6= 0. In this case, ap−1 6= 0, and we canwrite ap−1 = tlu, where l ∈ Z, u ∈ kv, vt(u) = 0, and vt denotes the valuationcorresponding to t. If l ≥ 0 then

vt(app−1 + dp−1ap−1/dtp−1) ≥ 0.

We note that for any a ∈ kv then vt(da/dt) ≥ vt(a)− 1, so

vt(dp−1ap−1/dtp−1) ≥ l − (p− 1).

If l ≤ −2 then

vt(dp−1ap−1/dtp−1) ≥ l − (p− 1) > pl = vt(a

pp−1),

hencevt(a

pp−1 + dp−1ap−1/dtp−1) = pl < −p = vt(ct

−p).

So, by comparing both sides of (5) we find that l = −1. By using Leibnitz’sFormula and Wilson’s Theorem again, we obtain

(6) ct−p = t−pup − t−pu +p−2∑i=0

Cip−1(−1)ii!t−(i+1)dp−1−iu/dtp−1−i.

Let u = u0 + tu1, u0 ∈ Fq, u1 ∈ Fq[[t]]. By comparing the coefficients oft−p in both sides of (6), we get c = up

0 − u0. Conversely, if c = up − u, forsome u ∈ Fq, then ct−1 = P (0, . . . , 0, ut−1).

From (1), (2), (4) we have |H1(kv, G)| = | ker ϕ|, where ϕ is the homo-morphism: (Fq, +) → (Fq, +), u 7→ up−u. It is clear that | ker ϕ| = |Fp| = p.Hence |H1(kv, G)| = p.

b) By Lemma 4.1, we have [k1/p : k] = p. By [Oe], Ch. VI, Sec.5, G isisomorphic to the quotient of the Weil restriction

∏k1/p/k

Gm by Gm, and from

[Oe, Chap. VI, Sec. 5.4, Proposition 2], it follows that G has Tamagawanumber τG = p. Then, by using the Hasse principle for the Brauer group of

19

global fields (see, e. g. [We]) one checks that III(G) = 1.Now, assume that H1(k,G) is finite. Then H1(kv, G) is trivial for almost allv, since

H1(k,G) →∏v∈F

H1(kv, G)

is surjective for any finite set F of discrete valuations ([TT]). Let

γ : H1(k,G) →∏v

H1(kv, G))

be the localization map and let S be the finite set of v such that H1(kv, G) 6= 0.The map H1(k, G) →

∏v

H1(kv, G)) is thus surjective, so by [Oe], Ch. IV,

Corollaire 3.4, we have the following formula for the Tamagawa number τG

of G :

τG =|III(G)||Coker γ|

=1

|Coker γ|≤ 1.

This is contradicting to the fact that τG = p. Therefore H1(k,G) is infinite.

c) Using the same argument as in part a), we get |H1(kv, G)| = | ker ϕ|,where ϕ is the homomorphism: (Fq, +) → (Fq, +), u 7→ up + u. Note thatker ϕ is a normal subgroup of Fq and has at most p elements, then | ker ϕ| = 1or p. If | ker ϕ| = p then the polynomial Xp−1 + 1 can be split into linearfactors over Fq[X] and Fq contains a (p− 1)th-root of −1. Conversely, if Fq

contains a (p − 1)th-root α of −1 then α, 2α, . . . , (p − 1)α are all roots ofXp−1 +1 and | ker ϕ| = p. It is clear that Fq contains a (p−1)th-root of −1 ifand only if Fq contains a primitive 2(p− 1)-root of unity, ζ2(p−1) . Since Fpn

is the splitting field of Xpn −X = 0 over Fp, Fpn contains ζ2(p−1) if and only

if ζpn−12(p−1) = 1, i.e., 2(p− 1) divides pn − 1, that is, if and only if n is even.

Further, we give another example of a wound unipotent group definedover Fp(t), which , when passing to Fq((t)), q = pn, has trivial Galois coho-mology group, for any n (compare with Proposition 4.3, c)).

4.4. Proposition. Let kv = Fq((t)), where q = 3n, and let G be theFp(t)-subgroup of G3

a defined as follows

G = {(x1, x2, x3)|x31 + tx3

2 + t−1x33 + x3 = 0}.

20

Then H1(kv, G) = 0.

Proof. Let P (x1, x2, x3) = x31 + tx3

2 + t−1x3 + x3. By an analogous argu-ment as in previous proposition, we have H1(kv, G) ' kv/P (kv × kv × kv).Hence, to prove the triviality of H1(kv, G), we need only prove the following:

(1) Fq[[t]] ⊂ P (Fq[[t]] × Fq[[t]] × Fq[[t]]), i.e.∑i≥0

citi ∈ P (Fq[[t]] × Fq[[t]] ×

Fq[[t]]), for all ci ∈ Fq,

(2) ct−i ∈ P (kv × kv × kv), for all c ∈ Fq, i ≥ 1.

Now, we prove (1). By differentiating both sides of the following equationwith respect to t twice∑

i≥0

citi = x3

1 + tx32 + t−1x3

3 + x3,

we get

(??)∑i≥0

2c3i+2t3i = 2t−3x3

3 + d2x3/dt2.

By writing x3 =∑i≥0

aiti, ai ∈ Fq we have

2t−3x33 + d2x3/dt2 = 2a3

0t−3 +

∑i≥0

(2a3i+1 + 2a3i+2)t

3i.

Comparing coefficients with the same power of t in (??) we get

a0 = 0, 2a3i+1 + 2a3i+2 = 2c3i+2, ∀i ≥ 0.

By setting a3i = a3i+1 = 0, a3i+2 = −a3i+1 + c3i+2,∀i ≥ 0, we get a solution

for (??). Since

d

dt

∑i≥0

(c3i+1t3i + 2c3i+2t

3i+1)− (−t−2x33 + dx3/dt)

= 0,

there exists x2 ∈ Fq[[t]] such that∑i≥0

(c3i+1t3i + 2c3i+2t

3i+1) = x32 − t−2x3

3 + dx3/dt.

21

Using this argument again, we have∑i≥0

citi = P (x1, x2, x3), for some x1, x2, x3 ∈

Fq[[t]].Finally, we prove (2) by induction on i. Namely, given any c ∈ Fq,

then c = β3, β = α3, α, β ∈ Fq. We have ct−1 = P (−α, 0, β), and ct−2 =P (0, βt−1, 0). So (2) is true for i = 1, 2. And it is easy to check thatct−3i = P (βt−i, 0, 0), ct−3i+1 = P (0, βt−i, 0), and ct−3i+2 = P (a, b, c+βt−i+1),where a, b, c ∈ K, satisfy −βt−i = P (a, b, c) (the existence of a, b, c are en-sured by the induction hypothesis).

4.5. Now we give a sufficient condition for the Galois cohomology groupsof unipotent commutative groups over non-perfect field to be infinite. Then,using this, we prove that the only connected smooth unipotent group of di-mension 1 with finite Galois cohomology group over global field is Ga. Beforestating and proving this condition, we need the following lemma. Recall thatif P is a p-polynomial in r variables T1, ..., Tr, with

P =∑

1≤i≤r

∑0≤j≤mi

cijTpj

i , where cimi6= 0,∀i,

then the principal part Pprinc of P is defined by

Pprinc =∑

1≤i≤r

cimiT pmi

i .

Lemma. Let k be a non-perfect field of characteristic p, v a non-trivial dis-crete valuation of k. Let P be a separable p-polynomial in r variables with

coefficients in k. Let Pprinc =r∑

i=1

ciTpmi

i be the principal part of P . Assume

that for all (a1, . . . , ar) ∈ kr, v(ci) + pmiv(ai) are all distinct whenever theyare finite. Then there exists a constant C0 depending only on P, such that ifa = P (a1, . . . , ar) and v(a) ≤ C0 then v(a) = v(ci) + pmiv(ai), for some i.

Proof. We process by induction on r. By assumption, P is a non-zero p-polynomial. First let r = 1, P (T ) = c0T + · · · + cm−1T

pm−1+ cmT pm

. Weset

A = infi,j{v(ci)− v(cj)

pj − pi| 0 ≤ i, j ≤ m, i 6= j},

B = inf {Api + v(ci) | 0 ≤ i ≤ m} − 1,

22

and pick any C0 with C0 < B. Now assume that a = P (a1) such thatv(a) = v(P (a1) ≤ C0 (a1 ∈ k). Let i0 be such that

v(ci0api0

1 ) = inf {v(ciapi

1 ) | 0 ≤ i ≤ m}.

Then we have C0 ≥ v(a) = v(P (a1)) ≥ v(ci0api0

1 ) = v(ci0) + pi0v(a1). Hence

(C0 − v(ci0))/pi0 ≥ v(a1).

By the choice of C0, B, and A we have

v(a1) < (B − v(ci0))/pi0 < (Api0 + v(ci0)− v(ci0))/p

i0 = A.

Hence by the definition of A

v(a1) <v(ci)− v(cm)

pm − pi, ∀i < m,

or, equivalently,

v(ciapi

1 ) = v(ci) + piv(a1) > v(cm) + pmv(a1) = v(cmapm

1 ), ∀i < r.

Therefore v(a) = v(cm) + pmv(a1) as required.Now we assume r > 1 and that the assertion of the lemma holds true

for all integers less than r. By induction hypothesis, for any l with 1 ≤l < r, there exist constants Bl satisfying the lemma for the case r = l.

Any monomial of P (T1, . . . , Tr) − Pprinc(T1, . . . , Tr) is of the form λT pmj−s

j ,λ ∈ k∗, 1 ≤ j ≤ r, s ≥ 1, and for such a monomial, we set

aλ,s,j =v(λ)− v(cj)

pmj − pmj−s,

and set

C3 =[

infλ,j,s,i

{(v(λ) + pmj−saλ,j,s − v(ci))p−mi}

]− 1.

Let

C2 =[

inf1≤i,j≤r

{(v(ci) + pmiC3 − v(cj))p−mj}

],

C1 = infi,j{v(cij) + pmijC2},

C0 = inf{C1, B1, . . . , Br−1}.

23

Assume that a = P (a1, . . . , ar), ai ∈ k and v(a) ≤ C0. If there exists i suchthat ai = 0 then |{i | ai 6= 0}| < r and instead of P we may consider thepolynomial P = P (T1, . . . , Ti−1, 0, Ti+1, . . . , Tr) in r− 1 variables and use theinduction hypothesis. So, we assume that ai 6= 0 for all i. Let

i0 = inf1≤i≤r

{i | v(ai) ≤ v(aj), for all j, 1 ≤ j ≤ r}.

Then

v(a) = v(P (a1, . . . , ar)) ≥ inf{v(cijapmij

i )} ≥ inf{v(cij) + pmijv(ai0)}.

By assumptions v(a) ≤ C0 ≤ C1, so we have v(ai0) ≤ C2 by definition of C1.Since v(ci) + pmiv(ai) are pairwise distinct, there exists unique i1 such that

v(ci1) + pmi1v(ai1) = inf1≤j≤r

{v(cj) + pmjv(aj)}.

Since

v(ci1) + pmi1v(ai1) ≤ v(ci0) + pmi0v(ai0) ≤ v(ci0) + pmi0C2,

we have v(ai1) ≤ C3, since otherwise, we would have

v(ci1) + pmi1v(ai1) > v(ci1) + pmi1C3 ≥ v(ci0) + pmi0C2,

which contradicts the above inequalities. Now, we show that

v(P (a1, . . . , ar)) = v(ci1) + pmi1v(ai1).

This follows from the following two facts:

(i) For any monomial λT pmj−s

j of P (T1, . . . , Tr)−Pprinc(T1, . . . , Tr), appearingin P (T1, ..., Tr), λ ∈ k, 1 ≤ j ≤ r, s ≥ 1, if v(aj) < aλ,j,s then by the definitionof aλ,s,j and i1, we have

v(λapmj−s

j ) = v(λ) + pmj−sv(aj) > v(cj) + pmjv(aj) ≥ v(ci1) + pmi1v(ai1).

Also, if v(aj) ≥ aλ,j,s then again by the definition of aλ,s,j and C3, we have

v(λapmj−s

j ) ≥ v(λ) + pmj−saλ,j,s > v(ci1) + pmi1C3 ≥ v(ci1) + pmi1v(ai1),

since v(ai1) ≤ C3. Thus, we have for all j

v(λapmj−s

j ) ≥ v(ci1) + pmi1v(ai1).

24

Hence

v(P (a1, ..., ar)− Pprinc(a1, ..., ar)) ≥ v(ci1) + pmi1v(ai1).

(ii) For j 6= i1, by the uniqueness of i1, we have

v(cjapmj

j ) > v(ci1) + pmi1v(ai1),

sov(Pprinc(a1, ..., ar)) = v(ci1a

pmi1

i1 +∑j 6=i1

cjapmj

j ) > v(ci1ap

mi1

i1 ).

Now from above it follows that

v(a) = v(P (a1, ..., ar))

= v(Pprinc(a1, ..., ar) + (P (a1, ..., ar)− Pprinc(a1, ..., ar))

= v((ci1ap

mi1

i1 +∑j 6=i1

cjapmj

j ) + (P (a1, ..., ar)− Pprinc(a1, ..., ar)))

= v(ci1ap

mi1

i1 ) = v(ci1) + pmi1v(ai1).

The proof of the lemma is completed.

4.6. Proposition. Let k be a non-perfect field of characteristic p, v anon-trivial discrete valuation of k. Let G be a commutative unipotent group

defined by a separable p-polynomial P in r variables, with Pprinc =r∑

i=1

ciTpmi

i ,

ci ∈ k∗. Let m = min{m1, . . . ,mr} and assume that v(c1), . . . , v(cr) are alldistinct modulo pm and r < pm. Then H1(k,G) is infinite.

Proof. Since v(c1), . . . , v(cr) are all distinct modulo pm, then for any tu-ple (a1, . . . , ar) ∈ k∗r, the values v(ci) + pmiv(ai), 1 ≤ i ≤ r, are alwayspairwise distinct. Then all assumptions in the above lemma are satisfied,so there exists C0 as in the lemma. Since r < pm then there exists l in{0, . . . , pm − 1} such that v(ci) 6≡ l(mod pm), for all i ∈ {1, .., r}. Weclaim that for all a with v(a) ≤ C0, and v(a) ≡ l (mod pm), then a isnot in P (kr). For, assume the contrary that a = P (a1, . . . , ar), ai ∈ k.By the lemma above, there is i such that v(a) = v(ci) + pmiv(ai). But

25

this is contradicting to the fact that v(ci) 6≡ l (mod p), hence the claimfollows. We can choose a sequence {bn}n, bn ∈ k for all n ≥ 1, suchthat C0 > v(b1) > v(b2) > · · · > v(bn) ≡ l (mod pm), for all n. Thenv(bn − bn+m) = v(bn+m) ≡ l (mod pm), for all m,n ≥ 1. Therefore by theclaim above, for all m, n ≥ 1 then bn − bn+m 6∈ P (kr), so all bn are distinctin k/P (kr). Therefore H1(k,G) is infinite as required.

Obvious examples (cf. 4.2, 4.3) show that one cannot relax the condition”r < pm” to ”r ≤ pm”, or relax the condition that the valuations v(ci) areall distinct (mod. pm). As application of the above result, we show that theflat cohomology in dimension 1 of one-dimensional non-split unipotent groupover non-perfect fields are infinite in general. First we consider the one-dimensional flat cohomology of elementary finite unipotent group schemes.In the simplest case of Z/pZ, the finiteness result dated back to Shatz [Sh].

4.7. Proposition. Let k be a non-perfect field of characteristic p > 0.a) Denote by αn the affine k-group scheme corresponding to k[T ]/(T n). Forany r ≥ 1, let G be a k-form of αpr or (Fp)

r. Then H1fl(k,G) is infinite.

b) Let G be a non-trivial unipotent k-group scheme of dimension 0. ThenH1

fl(k,G) is infinite.

Proof. a) In our situation, there exists a descending central series for G

G = G0 > G1 > · · · > Gn−1 > Gn = {0}

with successive quotients Gi/Gi+1 isomorphic to a k-form of (αp)r or (Fp)

r

(see [DG], or Raynaud [SGA 3, Expose XVII]). By making use of the factthat all groups Gi involved are commutative and H2

fl(k,Gi) = 0 (see e. g.[DG], Chap. III, §5, Corol. 5.8 , and [Se1], Chap. II, §2), it is clear thatit suffices to prove the proposition in the case r = 1. We can find and fixonce for all a non-trivial discrete valuation v of k. It is well known (see e.g.[DG], Chap. III, Raynaud [SGA 3], Expose XVII), that αp has no non-trivialk-forms and H1

fl(k, αp) ' k+/k+p. For any element a in k∗ with v(a) 6≡ 0(mod p), it is clear that a is not in kp. We may construct an infinite sequence{ai}, ai ∈ k, such that v(a1) < v(a2) < · · · < v(an) and v(an) 6≡ 0 (mod p)for all n. Then obviously ai (mod kp) are pairwise distinct, hence the abeliangroup k+/k+p (i.e. H1

fl(k, αp)) is infinite.

26

Now let G be a k-form of (Fp)r. Then we know that G can be embedded

as a closed k-subgroup of exponent p into Ga. The exact sequence

0 → G → Gaf→ Ga → 0

allows us to consider G as the kernel (hence also as zero set) of a separable p-polynomial f = f(T ) ∈ k[T ]. Since G has pr elements, we have deg(f) ≥ pr.Hence f(T ) = b0T +· · ·+bsT

ps, bs 6= 0, s ≥ r. Up to a k-isomorphism, we may

scale off bs, so we may assume that bs = 1. Then H1(k, G) ' k/f(k), wheref(x) = b0x + · · ·+ bs−1x

ps−1+ xps

Now, consider the constant C0 dependingonly on the p-polynomial f(T ) (with the only leading coefficient c = 1) asin Lemma 4.5. For z in k such that v(z) < C0 and v(z) 6≡ 0 (mod p),we see, according to this lemma, that z is not in f(T ). Again as above wemay choose an infinite sequence {ai} of elements from k such that ai 6∈ f(k),C0 > v(a1) > v(a2) > · · · > v(ai) > · · · for all i, and from this we concludethat H1(k, G) is infinite.

b) The proof is again by ”devisage”. First we assume that G is commu-tative. By Raynaud [SGA 3, Expose XVII], there is a composition series forG

G = G0 > G1 > · · · > Gr > Gr+1 > · · · > Gm = {0},

where each factor Gi/Gi+1 is a k-form of (Fp)r (resp. of (αp)

r) for 0 ≤ i ≤r − 1 (resp. for r ≤ m− 1). The rest follows from the proof of part a).

Now assume that G is not necessarily commutative and let C(G) denotethe center of G. We use induction on the length s = s(G) of ascending centralseries of G, which is finite since G is nilpotent. If s = 1, then G is commu-tative and we are back to the above case. Let s(G) = m > 1 and assumethat the assertion holds for all 1 ≤ s < m. We have s(G/C(G)) < s(G).Since G/C(G)) 6= {1} and has dimension 0, and since H2

fl(k, C(G)) = 0, theinduction hypothesis implies that H1

fl(k, G/C(G)) (and hence also H1fl(k, G))

is infinite.

For unipotent groups of positive dimension we have the following result.

4.8. Theorem. Let k be a non-perfect field of characteristic p. Let G be anon-trivial (not necessarily smooth) commutative unipotent k-group schemeof dimension ≤ 1, the connected component G◦ of which is not k-isomorphic

27

to an extension of Ga by an infinitesimal k-group scheme. Assume in addi-tion that if p = 2 then G◦ is neither k-isomorphic to an extension of Ga, norof the subgroup defined as in Proposition 4.2, b), by an infinitesimal groupscheme. Then the flat cohomology H1

fl(k,G) is infinite. In particular, for anyglobal field k of positive characteristic and for any non-trivial commutative k-unipotent group scheme G of dimension ≤ 1 with G◦ not k-isomorphic to anextension of Ga, by an infinitesimal k-group, the cohomology set H1

fl(k,G)is infinite.

Proof. I) First, we assume that G is commutative, smooth (i.e. absolutelyreduced) and connected.

If G has dimension 0, the assertion follows from Proposition 4.7.Assume that dim (G) =1. Then by [Ru], G is a k-form of Ga, and G is

k-isomorphic to a k-subgroup of Ga ×Ga given by

{(x, y)|ypn

= x + a1xp + · · ·+ anx

pm},

where a1, . . . , am are elements of k not all in kp. It is well known (see e.g. [Mi1]) that if G is reduced then H1

fl(k,G) ' H1(k,G). Let Pn(x, y) =x + a1x

p + · · ·+ amxpm − ypn, am 6= 0, then by an analogous argument as in

Proposition 4.2 we have H1(k,G) ' k/Pn(k2).

a) We consider the case p > 2. Evidently, Pn(k2) ⊂ P (k2), where P := P1.Therefore, we need only consider the case n = 1. Let v be a non-trivialdiscrete valuation with a prime element π, kv the completion of k respect tov. Then we know (cf. [TT]) that the localization map H1(k,G) → H1(kv, G)is surjective. Therefore, it suffices to consider the local case, i.e, to show thatH1(kv, G) is infinite. In the sequel we identify kv with the field k(v)[[π]]. Ifam ∈ kp

v then by changing the variables

y′ = y − αxpm−1

, x′ = x,

where am = αp, α ∈ kv, G is isomorphic to the group given by

{(x′, y′)|y′p = x′ + a1x′p + · · ·+ am−1x

′pm−1

}.

Hence, by repeating this argument (if necessary), we may assume that am

is not in kpv , with m ≥ 1, since by assumption, G 6' Ga. Assume that

v(am) ≡ 0 (mod p). We recall the following result from [Ru], Proposition

28

2.3. Denote by F the Frobenius morphism, A = k[F ] the (non-commutative)ring of polynomials in F subject to the relation Fa = apF , for all a ∈ k.Then A ' Endk(Ga,Ga), the ring of k-endomorphisms of Ga. Every k-form G of the k-group Ga is then defined by the tuple (F n, τ), where τ =a0 + a1F + · · · + amFm ∈ A, a0 6= 0 (a separable endomorphism), and it isk-isomorphic to the k-subgroup of Ga ×Ga defined by the equation

ypn

= a0 + a1xp + · · ·+ amxpm

.

By [Ru, Prop. 2.3], the two k-forms (F n, τ), (F n1 , τ1), with n1 ≤ n, τ1 =c0+c1F + · · ·+cm1F

m1 , are k-isomorphic if and only if there exist a separableendomorphism ρ = b0 + b1F + · · ·+ brF

r ∈ A, σ ∈ A, and c ∈ k∗ such that

(∑

0≤j≤m1

cpn−n1

j F j)c = (∑

i

bpn

i F i)τ + F nσ.

Using this we may choose a k-form of Ga defined by the equation yp =a′0 + a′1x

p + · · · + a′nxpn

, which is still k-isomorphic to G, such that v(a′n) 6≡0 (mod p). Now, all conditions in Proposition 4.6 are satisfied. Namely,

we have Pprinc = a′mx′pm

− y′p, r = 2 < p = pinf{1,m}, and v(a′m), v(−1) aredistinct modulo p. Therefore, by Proposition 4.6, H1(kv, G) is infinite.

b) Now we consider the case when p = 2.b1) Assume first that G is not isomorphic to the subgroup of Ga×Ga definedby the equation

{(x, y)|y2n

= x + a1x2},

where n ≥ 2, a1 6∈ k2. By the same argument as in the case p > 2, it sufficesto consider the local case, i.e., over kv, n = 1, m ≥ 2, and by applying [Ru,Prop. 2.3] if necessary,, we may assume that v(am) ≡ 1 (mod 2). But thenv(w2) = 2v(w) and v(am)+2mv(u) are always distinct whenever (u, w) ∈ k∗2.We choose an odd number l (mod 4) such that v(am) 6≡ l (mod 4). ByLemma 4.5, there exists a constant C0 such that if a = P (u, w), v(a) ≤ C0

then v(a) = v(w2) or v(a) = v(am) + 2mv(u). We claim that if a ∈ kv sat-isfies the conditions v(a) ≡ l (mod 4) and v(a) ≤ C0 then a 6∈ P (k2). Infact, assume that a = P (u, w), u, w ∈ k. Since v(a) ≤ C0, v(a) = v(w2) orv(am) + 2mv(u). But this contradicts to the fact that v(a) ≡ l (mod 4).We conclude as in the proof of Proposition 4.6 that H1(k, G) is infinite.

b2) We now assume that p = 2 and G is isomorphic to a subgroup of the

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above Russel’s form. It is sufficient to treat the case n = 1, m = 2. By mak-ing use of [Ru, Prop. 2.3], we may again assume that v(a1) ≡ 1 (mod 2).Then v(a1) ≡ 1 or 3 (mod 4). So, v(w4) and v(a1) + 2v(u) are alwaysdistinct whenever (u, w) ∈ k∗2. We use an analogous argument as in previouscase and we see that H1(k,G) is infinite.

II) We now assume that G is commutative, connected and non-smooth. Letαpr be the k-group scheme represented by Spec k[T ]/(T pr

). If dim G = 0,then G is a k-form of αpr for some r, thus isomorphic to the latter (here weused the fact that r-multiple extension of αp by itself is isomorphic to αpr

(see Raynaud [SGA 3, Exp. XVII]), and by Proposition 4.7, it has infiniteflat cohomology.

Now we assume that dim G = 1. Denote by F the Frobenius morphismon G, G(p) the k-group scheme obtained from G via the base change k → kp.For any natural number n, by considering F n we obtain G(pn) and we denote

F nG := Ker (F n) : G → G(pn). Then it is well-known (see Gabriel [SGA 3,Exp. VIA]) that there is a natural number n0 such that G/F nG is smoothfor all n ≥ n0. Let r be the smallest such a number. Then r ≥ 1 since G isnon-smooth, and we have the following composition series of characteristicsubgroups of G

G ≥ G◦ ≥ F rG ≥ · · · ≥ F G ≥ (0).

Here G◦ denotes the connected component of G, and each factor F s+1G/F sG,1 ≤ s ≤ r − 1 is a radicial (infinitesimal) k-subgroup scheme of height 1(loc.cit). Therefore we have an exact sequence

(?) 0 → αpr → G → Gred → 0,

where Gred is reduced. Then Gred is a k-form of Ga. From the exact se-quence (?), we derive the exact sequence of flat cohomology

H1fl(k,G)

f→ H1fl(k,Gred) → H2

fl(k, αpr).

By [DG], Chap. III, §5, Corol. 5.8, H2fl(k, αpr) = 0, so f is surjective. By

assumption, Gred 6' Ga, and if p = 2, Gred is not k-isomorphic to the groupdefined in Proposition 4.2, b), also. So by previous part, H1

fl(k,Gred) 'H1(k,Gred) is infinite, and it follows that the same is true for H1

fl(k,G).

30

III) Next we assume that G is commutative, but not connected (and notnecessarily smooth) so G/G0 6= {1} and consider the following exact se-quence

1 → G0 → G → G/G0 → 1,

where G0 denotes the connected component of G. Then we have dim(G)= dim(G0) = 1. Also, since G is commutative we have the following exactsequence of flat cohomology

H1fl(k,G◦) → H1

fl(k, G)β→ H1

fl(k,G/G◦) → H2fl(k,G◦).

Since G◦ is a commutative unipotent group scheme, from the exact sequence(like (∗) above)

0 → αpr → G◦ → G◦red → 0,

and from the triviality of H2fl for αpr and G◦

red, it follows that H2fl(k, G◦) = 0,

so β is surjective. It then suffices to show that H1fl(k,G/G0) is infinite. It is

known that H := G/G0 is a finite etale unipotent k-group scheme, hence ithas a composition series

H = H0 > H1 > · · · > Ht = {0}

with successive factors Hi/Hi+1 isomorphic to a k-form of (Fp)r for some r.

By Proposition 4.7, H1fl(k,H) is infinite as well, hence so is H1

fl(k, G).

IV) Now assume that k is a global function field of characteristic p. Werefer to Proposition 4.2, b), Proposition 4.7, and the previous parts for therest of the proof.

(Notice that, by using [DG], III, Section 6, one can show that the con-dition that G◦ not to be an extension of Ga by an infinitesimal unipotentk-group can be relaxed to the condition that G◦ 6' Ga, which is not essen-tial.)

4.9. Corollary. Let k be a non-perfect field of characteristic p > 0. Then,for all r, there exists a k-form of Gr

a such that H1(k, G) is infinite.

31

Proof. It is well known that for any non-perfect field k, there are manydiscrete valuations of k. For example, if p =char.k, t a transcendental ele-ment of k over Fq, then one may associate a nontrivial discrete valuation vt

associated to t (see e.g. [Bou], Chap. VI). Then by using Proposition 4.6,we may choose a k-form G of Gr

a with infinite H1(k,G).

It is quite possible that for any k-split unipotent group G over non-perfectfield k, there is a k-form G′ of G which has infinite cohomology. In [Ro2],Rosenlicht gave the following example of a wound noncommutative unipo-tent group G over a non-perfect field of characteristic 3. We are interested incomputing its Galois cohomology in degree 1 and show that there are manycases where the cohomology is infinite.

4.10. Example. Let k be a non-perfect field of characteristic p > 2. Let abe an element of k such that 1, a, a2 are linearly independent over kp. Let

Γ = {(c1, c2) ∈ G2a|c

p1 + acp

2 + ac1 = 0},

G1 = {(x, y) ∈ G2a|yp − y = axp},

G2 = {(x1, x2, x3) ∈ G3a|x

p1 + axp

2 + a−1xp3 + x3 = 0}.

For γ = (c1, c2) ∈ Γ, let

ϕγ(x, y) = ϕ(c1,c2)(x, y) = (c1x− c2y, c2x,−c1y).

This yields an homomorphism ϕ : Γ → Hom(G1, G2), γ 7→ ϕγ. On G :=Γ×G1 ×G2, we define the multiplication as follow

(γ, g1, g2)(γ′, g′1, g

′2) = (γ + γ′, g1 + g′1, g2 + g′2 + ϕγ(g

′1)).

Then G is a wound noncommutative unipotent group. By a direct checking,G2 is central in G and from the exact sequence 1 → G2 → G → G/G2 → 1,we have the following exact sequence (of pointed sets)

H1(k, G2) → H1(k,G) → H1(k, Γ×G1) → 1.

When k = Fq((t)), q = 3n, a = t, then by Proposition 4.5, H1(k, G2) = 0.Then H1(k,G) ' H1(k, Γ×G1), and H1(k,G) has a group structure via this

32

bijection. In general, by Theorem 4.8, H1(k,G1) is infinite (since p > 2),hence so is H1(k, G).

4.11. Remark. If p = 2 (resp. p = 3) then Proposition 4.2 (resp. Proposi-tion 4. 3) shows that there are examples of a global field k and commutativeunipotent k-groups G of dimension 1 (resp. 2) where H1(kv, G) is finite , butnon-trivial.

Theorem. Let k be a non-perfect field of charateristic p > 0.1) The one-dimensional Galois cohomology H1(k, G) of any connected smoothunipotent non-k-split k-group G is non-trivial (resp. infinite) if and only thesame is true for connected smooth commutative unipotent k-wound groupsof exponent p. In particular, over global function fields k, one-dimensionalGalois cohomology of non-split smooth unipotent k-groups of dimension ≤ 1are always infinite.2) The study of the triviality (resp. finiteness) of H1

fl(k, G) for unipotentk-group schemes G can be reduced in a canonical way to that for connectedsmooth commutative k-wound unipotent groups of exponent p.

Proof. 1) The ”Only if” part is trivial.

(”If” part.) We assume that the assertion of the theorem holds for one-dimensional smooth unipotent groups. Let G be any smooth unipotent non-split k-group of dimension n. We use induction on n. If n = 1, there isnothing to prove, and we assume that n > 1 and the assertion holds for allm < n. Denote by Gs the k-split part of G. We consider two cases.

a) Gs 6= 1. Then G/Gs is unipotent and k-wound as is well-known, andit has dimension strictly less than n. We consider two subcases.

a1) If Gs is commutative, we consider the exact sequence of k-groups

1 → Gs → G → G/Gs → 1,

and the cohomology sequence derived from this

0 → H1(k, G)π→ H1(k,G/Gs) → H2(k,Gs) = 0,

thus π is surjective. By inductive assumption, H1(k, G/Gs) is non-trivial(resp. infinite), so the same is true for H1(k,G).

33

a2) Assume that Gs is not commutative. Denote by

Gs = G0 > G1 > G2 > · · · > Gl = {1}

the derived series of Gs, i.e., Gi+1 = [Gi, Gi], for all 0 ≤ i ≤ l − 1, and Gl−1

is commutative and non-trivial, hence it has positive dimension. One checksthat Gl−1 is a normal subgroup of G and the assertion of the theorem followsfrom the exact sequence

1 → Gl−1 → G → G/Gl−1 → 1

by combining with the above arguments.

b) Gs = 1, i.e., G is k-wound. If G is commutative and of exponent p,there is nothing to prove. Assume that G is not of this type. Then it is well-known by Tits theory again ([Oe], [Ti]), that there exists a central connectedcommutative k-subgroup K of G of exponent p with dimK > 0, such thatG/K is k-wound. The assertion now follows by using the exact sequence

1 → K → G → G/K → 1

and the induction hypothesis.

2) We first use induction on the length of the ascending central series of G.To reduce the length, we consider the center C(G) of G, which is non-trivial,since G is nilpotent. The exact sequence 1 → C(G) → G → G/C(G) → 1will do the job. Therefore we are reduced to the case G is commutative.The exact sequence 1 → G◦ → G → G/G◦ → 1, combined with Proposition4.7 reduces the situation further to the case G is connected. Again by usingProposition 4.7, we are reduced to the case G is smooth and connected. Nowthe rests follows from the last part of the proof of 1).

5 Some local - global principles and applica-

tions

We derive from Theorem 3.1 and its proof and from [Oe] the following

34

5.1. Theorem. (Local - global principle). Let k be a global field and let Gbe a smooth group defined over k with finite Shafarevich - Tate group.a) If G satisfies the condition (∗) of Section 3, then H1(k, G) is trivial if andonly if H1(kv, G) is trivial for all valuations v of k and for some (hence any)embedding of G into a k-group H = SLn such that with natural action of Hon H/G, the orbits of H(k) are closed in (H/G)(k) in the A(S)-topology of(H/G)(k) for any finite set S of valuations of k.b) If G is unipotent, then the assertion a) also holds for a k-split unipotentgroup H.

Proof. In fact we prove a slightly stronger assertion as follows.

(•) If H1(k, G) is trivial, then for any valuation v of k, H1(kv, G) is alsotrivial. Conversely, if H1(kv, G) is finite for all v and it is trivial for almostall v, and if the above embedding condition holds, then H1(k,G) is also trivial.

Proof of (•). The first part of the statement is trivial in the case a) andfollows from the proof given above in Section 3 for the case b). For thesecond one, recall that for smooth k-group G, the Tate - Shafarevich group

III(G) := Ker (H1(k, G) →∏v

H1(kv, G))

is finite according to [Oe], Chap. IV, Prop. 2.6. It follows easy from Corollary3.4 that for any finite set S of valuations of k, the localization map

ϕS : H1(k,G) →∏v∈S

H1(kv, G)

is surjective (in fact a more general result holds true, see [TT]). We takeS := { v | H1(kv, G) 6= 0} (which may be empty); then we have an exactsequence of abelian groups

1 → III(G) → H1(k, G) →∏v∈S

H1(kv, G) → 1.

From this and from the assumption it follows that H1(k, G) is finite as well,and by our theorem, H1(k,G) is trivial as required.

With above notation we have also the following

35

5.2. Corollary. a) If T is a smooth k-torus, K a finite radicial (i.e. purelyinseparable) extension of k of degree pn, and G = RK/k(TK), then the k-groupU := G/T has trivial (resp. finite) 1-Galois cohomology if and only if bothgroups H1(k, T )/pn and pnH2(k, T ) are trivial (resp. finite).b) For any smooth k-torus T and finite set S of non-equivalent valuations ofk, the localization map

pnH2(k, T ) →∐v 6∈S

pnH2(kv, T )

is surjective.c) If T, G and U are as in a), then for U the obstruction to weak approxima-tion in S is finite, i.e., the quotient A(S, U) =

∏v∈S

U(kv)/Cl(U(k)) is a finite

abelian group.

Here for an abelian group A and a natural number n we denote byA/n (resp. nA) the cokernel (resp. kernel) of the natural endomorphism

A×n→ A, x 7→ nx.

Proof. a) According to [Oe], Appendix 3, U is a smooth unipotent group,and by loc.cit, Chap. VI, Sec. 5.2, we have the following exact sequence

H1(k, T )pn

→ H1(k, T ) → H1(k, U) → H2(k, T )pn

→ H2(k, T ),

thus we have also an exact sequence

0 → H1(k, T )/pn → H1(k, U) → pnH2(k, T ) → 0.

Hence H1(k, U) is trivial (resp. finite) if and only if both groups H1(k, T )/pn

and pnH2(k, T ) are trivial (resp. finite).

b) Follows from the exact sequence above and Corollary 3.4.

c) We prove a more general assertion. Namely

(#) if U is a k-unipotent group for which there is a separable dominantk-morphism f : V → U , where V is an open set in some affine space An

36

then the assertion of the corollary holds.

Indeed, by Implicit Function theorem (see [Se2, Part 2]), being separable,f defines open maps fv : V (kv) → U(kv) for all valuations v of k, accordingto implicit function theorem. Since V (k) is dense in

∏v∈S

V (kv), it follows

that Cl(f(V (k))) is open in∏v∈S

U(kv) and so is Cl(U(k)). Therefore A(S, U)

is discrete in its natural quotient topology. If U is k-wound then U(kv)is compact by [Oe], Chap. VI, Prop. 2.1. Therefore A(S, U) is finite asrequired. In general case, we have the following exact sequence

1 → Us → U → Q → 1,

where Us is the maximal k-split normal connected subgroup of U , Q = U/Us

(see [Ti], [Oe]). We apply Lemma 3.3 above to the following commutativediagram

Us(k) → U(k) → Q(k) → 1

α ↓ ↓ β ↓ γ′

Us(S) → U(S) → Q(S) → 1

where X(S) :=∏v∈S

X(kv), with X stands for U,Us, Q, and Us(k) is dense

in Us(S) due to weak approximation in the k-split group Us. We have ahomeomorphism

A(S, U) = U(S)/Cl(U(k))

' Q(S)/Cl(Q(k))

= A(S, Q).

Since Q is k-wound ([Ti], [Oe]), then for any valuation v, Q(kv) is compact([Oe], loc.cit.), thus A(S, Q) is also compact. Since A(S, U) is discrete (seeabove), it follows that all they are finite as required.

37

The group U = G/T satisfies our assumptions in (#) by [Oe], Chap. VI,hence c) holds.

5.3. We now consider some applications to local - global principles foruniversality of polynomials over global fields. We say that a polynomialF (T ) := F (T1, ..., Tn) ∈ k[T1, ..., Tn] is universal over k if it represents anyelement from k, i.e., the equation F (T ) = a has solution in kn for any a ∈ k.As an easy consequence of the Hasse - Minkowski theorem (local - globalprinciple) for quadratic forms we have the following (perhaps well-known)

Proposition. Let f = f(X1, ..., Xn) be a quadratic form in n variablesover a global field of characteristic p 6= 2. Then f is universal over k if andonly if f is so over all completions kv of k.

Proof. Assume that f is universal over k. It is clear that k∗2v is open in kv. Forany bv ∈ k∗v we pick b ∈ k∗ such that b is sufficiently close to bv in the v-adictopology, and so that bv = bc2

v, where cv ∈ k∗v . Since f(kn) = k by assump-tion, we see that there is a ∈ kn such that f(a) = b, hence f(cva) = bc2

v = bv,i.e., f is also universal over kv.

Conversely, assume that f is universal over kv for all v. Take any b ∈ k∗

and consider the quadratic form f ′ := f − bX2, where X is a new variable.Since f is universal over kv for all v, one sees that f represents b over kv forall v, i.e., the quadratic form f ′ is isotropic over kv, for all v. Now Hasse- Minkowski Theorem for quadratic forms over global fields (see e. g. [Sc])tells us that f ′ is also isotropic over k. Denote by (x1, ..., xn, x) ∈ kn+1 be anon-trivial zero of f ′. If x = 0, then (x1, ..., xn) is a non-trivial zero of f , sof is isotropic over k. Hence f is universal over k and f represents b over k.Otherwise, x 6= 0 and we may divide by x2 to see that again f represents bover k.

5.4. A polynomial F (T ) in n variables T = (T1, ..., Tn) with coefficients in afield k is called additive, if F (X + Y ) = F (X) + F (Y ) for any X, Y ∈ kn.Clearly, any algebraic group morphism from vector groups to the additivegroup Ga is an additive polynomial. It is well-known (see e. g. [Go], [W1,W2]) that additive polynomials play important role in the study of arith-metic of the global function fields. In particular, its universality plays alsosome role in the model-theoretic approach to the arithmetic of function fields(cf. e. g. [Ku]).

38

Theorem. Let F be a separable additive polynomial in n variables withcoefficients in a global field k.a) F is universal over k if and only if F is universal over kv for all valuationsv of k, and F (kn) is A(S)-closed in k for all finite set S of valuations of k .b) Assume that n=2. Then F is universal over K if and only if it is so overall kv.

Proof. a) By assumption, F can be regarded as a separable morphism ofalgebraic groups F : Gn

a → Ga. The kernel of F is then a commutativesmooth unipotent k-subgroup GF of Gn

a , and we have the following exactsequence of k-groups

1 → GF → Gna

F→ Ga → 1.

It is well-known that F is a p-polynomial, where p is the characteristic ofthe field k. Thus if p = 0, there is nothing to prove. Let p > 0. The exactsequence of Galois cohomology related to the exact sequence above gives usthe first Galois cohomology of GF as follows : For any extension field K/kwe have

H1(K, GF ) ' K/F (Kn).

Thus our additive polynomial F is universal over a field K if and only ifH1(K, GF ) = 1. Therefore, from Theorem 5.1 it follows that over a globalfield k, an additive separable polynomial is universal over k if and only if itis so over all completions kv of k and the condition regarding the closednessholds.b) With notation as in the proof of a), GF is a connected and smooth unipo-tent k-group of dimension 1. If H1(k,GF ) = 1, then from the surjectivity ofthe localization map we know that H1(kv, G) = 1 for all v.

Conversely, assume that H1(kv, GF ) = 1 for all v. Assume first that p 6= 2.Then it follows from the exact sequence

1 → III(GF ) → H1(k,GF ) →∏v

H1(kv, GF ) → 1

and the finiteness of III(GF ) ([Oe], Ch. IV) that H1(k,GF ) is also finite. Nowfrom Proposition 4.6 it follows H1(k,GF ) is either trivial or infinite, henceH1(k,GF ) = 1 as required. Therefore F is universal over k if and only if itis so over all kv.

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Let p = 2. If GF were not isomorphic to Ga then from Propositions 4.2 and4.6 it would follow that H1(k,GF ) is infinite, so GF ' Ga, and the proposi-tion follows.

5.5. Remarks. 1) The ”only if” part of Theorem 5.1 in fact, holds true forany field as it follows from a more general result in [TT].

2) One cannot replace the statement regarding the universality (i.e., therepresentation of all elements of k by F ) of the additive polynomial F con-sidered by the statement regarding the representation of an element of k byF . Namely, using [Oe], Chap. V, one can construct a p-polynomial F andan element a ∈ k such that the equation F (T1, ..., Tn) = a has solutions in kn

v

locally everywhere, but has no solutions in kn, i.e., the local - global principlefails in this case.

The structure of p-polynomials reflects deep arithmetic properties of thebase field. As an easy consequence of the above we have

5.6. Corollary. Let k be a non-perfect field of characteristic p > 0. Then,for any n, k has cyclic extensions of degree pn.

Proof. By Proposition 4.7, for any k-form G of Fp, the Galois cohomol-ogy H1(k,G) is infinite. In particular, there exists α ∈ k \ ℘(k), where℘(x) = xp − x is the Artin - Schreier polynomial. By a well-known theoremof Witt [Wi, Satz 13], for any n there exist cyclic extensions of k of degreepn.

5.7. Remark. P. Russell showed in [Ru], p. 538, that

(∗) if a field k of characteristic p > 0 does not have normal extensions ofdegree p, then for any p-polynomial f(T ) = a0T +a1T

p + · · ·+amT pm ∈ k[T ],with a0 6= 0, we have f(k) = k, and H1(k, G) = 0 for all G which are k-formsof Ga.

In other words, for the following statements

a) k has no normal extensions of degree p;

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b) every separable p-polynomial f(T ) = a0T + a1Tp + · · · + amT pm

(a0 6= 0)is universal, i.e., f(k) = k;

c) Every k-form G of Ga has trivial 1-Galois cohomology,

we have the following implications

a) ⇒ b) ⇒ c).

Equivalently, if the field k is such that there is a k-form G of Ga with non-trivial 1-cohomology, then k has a normal extension of degree p. (This canbe derived immediately from Corollary 5.6, since such a field k must benon-perfect, according to Rosenlicht’ Theorem, hence by Corollary 5.6 it hascyclic extensions of degree pn for any n.) There (in [Ru], p. 538) it wasalso mentioned that ”the author does not know whether the converse of thisstatement is true if k is not perfect.”

In other words, one may wonder if the following implications are true ifk is not perfect :

c) ⇒ b) ⇒ a).

Our result (Proposition 4.6) gives a clarification to this question. Namely,with the assumption on the non-perfectness of k, it is automatically truethat k has normal extensions of degree p, and that there are k-forms G of Ga

with non-trivial one-dimensional Galois cohomology. Therefore the aboveimplications also hold true.

Acknowledgements. We would like to thank T. Kambayashi, J. Oesterleand P. Russell for their attention and support via email correspondences.The first author would like to thank the Abdus Salam I. C. T. P. and I. H.E. S. for the support and excellent working condition while preparing thisversion of the paper.

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